# Thread: piecewise functions and continuity.

1. ## piecewise functions and continuity.

I know the answer to 1a is No, but i dont know why, just wanting to know the theory behind it and i dont know how to do 1b...any help would be appreciated. thanks

2. Originally Posted by b00yeah05

I know the answer to 1a is No, but i dont know why, just wanting to know the theory behind it and i dont know how to do 1b...any help would be appreciated. thanks
each piece of the function is a polynomial, so we know each piece is continuous on its own. however, we must be concerned with where the function breaks. this is at x = 3.

for 1a, k = 1 does not work because when we plug in k = 1 and x = 3 we end up with two different values for each piece, which obviously (why is this obvious?) makes the function not continuous.

for 1b. we need to find k so that the function is continuous. meaning, when x = 3, we want both functions to give the same value. how do you think we should proceed?

3. well the obvious reason would be that it is different values and hence it cant be continuous for 1a, and as for 1b i was thinking maybe solve them as a simultaneous equation?

4. Originally Posted by b00yeah05
[snip]
as for 1b i was thinking maybe solve them as a simultaneous equation?
Not quite ....

3^2 + 3k - 27 = -(3)^2 + 7(3) + k (and how did I get this?)

=> .......

5. you made x=3 and then solving...hence k = 24. at that value the function is continuous. am i correct??

6. Originally Posted by b00yeah05
you made x=3 and then solving...
yes. but the important step is equating the two formulas. this ensures that they are equal when we're done

7. i edited my post before..sorry..am i correcT?

8. Originally Posted by b00yeah05
i edited my post before..sorry..am i correcT?
no

9. what do i do after i get the co-efficient of k = 29?

10. Originally Posted by b00yeah05
what do i do after i get the co-efficient of k = 29?
that is wrong as well. remember, we are solving 9 + 3k - 37 = -9 + 21 + k for k

11. ah...i think you might have missed the negative sign there...it is negative x^2 in the second function so (-3)^2 = 9 which would cancel the 9's out?

12. Originally Posted by b00yeah05
ah...i think you might have missed the negative sign there...it is negative x^2 in the second function so (-3)^2 = 9 which would cancel the 9's out?
you have -x^2, so if you plug in x = 3 you have -(3)^2 = -9

13. i see...so thats where i was going wrong..so k=20

14. Originally Posted by b00yeah05
i see...so thats where i was going wrong..so k=20
yes

15. Originally Posted by b00yeah05
ah...i think you might have missed the negative sign there...it is negative x^2 in the second function so (-3)^2 = 9 which would cancel the 9's out?
Originally Posted by Jhevon
you have -x^2, so if you plug in x = 3 you have -(3)^2 = -9
*Ahem* b00yeah05 ........ which is why I wrote -(3)^2, NOT (-3)^2:

Originally Posted by mr fantastic
Not quite ....

3^2 + 3k - 27 = -(3)^2 + 7(3) + k (and how did I get this?)

=> .......
A flagon of frustration might have been saved if the original equation (above) that started this almost-death by a thousand cuts had been read and worked a little more ...... carefully?

No need to thank me for the above (apparently useless) post - the pleasure's in the giving.